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Kinia7

Rejestracja: 16 Jan 2012
Offline Ostatnio: Jan 15 2018 10:00
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Moje posty

W temacie: Oblicz wykładniki z danych liczb

15.01.2018 - 09:58

Np. d)

 

42,426407\ /^2\quad\to\quad\approx 1800=900\cd2=\(30\sq2\)^2\quad\to\quad 42,426407\approx30\sq2


W temacie: Stosunek promieni.

31.12.2017 - 23:36

a=BC=7\ \ \ \ \ b=AC=9\ \ \ \ \ c=AB\ \ \ \ \ \alpha=\angle BAC
\angle ABC=2\alpha\ \ \ \ \ \angle ACB=180^{\circ}-3\alpha
z tw. sinusów  \fr{7}{\sin\alpha}=\fr{9}{\sin2\alpha}=\fr{9}{2\sin\alpha\cos\alpha}\quad\to\quad \cos\alpha=\fr9{14}\quad\to\quad \sin\alpha=\sq{1-\cos^2\alpha}=\fr{\sq{115}}{14}
\sin3\alpha=\sin2\alpha\cos\alpha+\sin\alpha\cos2\alpha=\fr{16\sq{115}}{343}
z tw. sinusów  \fr{c}{\sin(180^{\circ}-3\alpha)}=\fr{a}{\sin\alpha}=2R\quad\to\quad c=\fr{7\sin3\alpha}{\sin\alpha}\ \ \ \ \ R=\fr{7}{2\sin\alpha}
\{P=\fr12ab\sin(180^{\circ}-3\alpha)=\fr{63}{2}\sin3\alpha\\P=r\cd\fr{a+b+c}{2}=r\cd\fr{16+\fr{7\sin3\alpha}{\sin\alpha}}{2}   \quad\to\quad r=\fr{63\sin3\alpha\sin\alpha}{16\sin\alpha+7\sin3\alpha}
\fr Rr=\fr{16\sin\alpha+7\sin3\alpha}{18\sin3\alpha\sin^2\alpha}=\fr{16\cd\fr{\sq{115}}{14}+7\cd\fr{16\sq{115}}{343}}{18\cd\fr{16\sq{115}}{343}\cd\fr{115}{196}}=\fr{343}{115}

W temacie: Wykaż - równanie trygonometryczne sin 7

31.12.2017 - 23:34

\sin47^{\circ}+\sin61^{\circ}-\(\sin11^{\circ}+\sin25^{\circ}\)=
=2\sin\fr{47^{\circ}+61^{\circ}}{2}\cos\fr{47^{\circ}-61^{\circ}}{2}-2\sin\fr{11^{\circ}+25^{\circ}}{2}\cos\fr{11^{\circ}-25^{\circ}}{2}=
=2\sin54^{\circ}\cos7^{\circ}-2\sin18^{\circ}\cos7^{\circ}=
=2\cos7^{\circ}\(\sin54^{\circ}-\sin18^{\circ}\)=
=2\cos7^{\circ}\cd2\sin\fr{54^{\circ}-18^{\circ}}{2}\cos\fr{54^{\circ}+18^{\circ}}{2}=
=4\cos7^{\circ}\sin18^{\circ}\cos36^{\circ}=
=4\cos7^{\circ}\sin18^{\circ}(1-2\sin^218^{\circ})=
=4\cos7^{\circ}\fr{\sq5-1}{4}\(1-2(\fr{\sq5-1}{4})^2\)=
 
1-2\left(\frac{\sqrt{5}-1}{4}\right)^2=1-\frac{\left(\sqrt{5}-1\right)^2}{8}=1-\frac{3-\sqrt{5}}{4}=\frac{1}{4}+\frac{\sqrt{5}}{4}
 
=4\cos7^{\circ}\fr{\sq5-1}{4}\cd\fr{\sq5+1}{4}=\cos7^{\circ}

W temacie: Tw: sin, cos i tw. o dwusiecznych

31.12.2017 - 23:33

BM=m\ \ \ \ BN=n\ \ \ \ MN=d\ \ \ \ \angle MBN=\beta
m=c\cos\beta\ \ \ \ \ n=a\cos\beta
\{P_{ABC}=\fr12ac\sin\beta \quad\to\quad ac=\fr{2P_{ABC}}{\sin\beta}\\P_{BMN}=\fr12mn\sin\beta=\fr12ac\cos^2\beta\sin\beta \quad\to\quad ac=\fr{2P_{BMN}}{\cos^2\beta\sin\beta}   \quad\to\quad \cos^2\beta=\fr{P_{BMN}}{P_{ABC}}
\sin^2\beta=1-\fr{P_{BMN}}{P_{ABC}}=\fr{P_{ABC}-P_{BMN}}{P_{ABC}} \quad\to\quad \sin\beta=\sq{\fr{P_{ABC}-P_{BMN}}{P_{ABC}}}
promień  r  okręgu opisanego na  \triangle BMN  z tw. sinusów  
 \fr{d}{\sin\beta}=2r \quad\to\quad r=\fr{d}{2\sin\beta}=\fr{d}{2\sq{\fr{P_{ABC}-P_{BMN}}{P_{ABC}}}}
\triangle BMN\approx\triangle ABC  w skali  k=\sq{\fr{P_{BMN}}{P_{ABC}}}
R=\fr rk=\fr{\fr{d}{2\sq{\fr{P_{ABC}-P_{BMN}}{P_{ABC}}}}}{\sq{\fr{P_{BMN}}{P_{ABC}}}}=\fr{d\cd P_{ABC}}{2\sq{P_{BMN}(P_{ABC}-P_{BMN})}}=\fr{2\cd18}{2\sq{2\sq2(18-2\sq2)}}=\fr{9(9\sq2+2)\sq{9\sq2-2}}{158}

W temacie: Prostopadłościan

31.12.2017 - 23:32

AB=a=1,\ BC=b=2,\ DS=h=3  - boki prostopadłościanu;  BD=p  - przekątna podstawy;  BS=q  - przekątna prostopadłościanu;  BM=x\angle ABS=\beta\ \ \ \angle DBS=\gamma
p^2=a^2+b^2=5\ \ \ \ \ \ \ q^2=a^2+b^2+h^2=14
z tw. kosinusów w  \triangle ABS\ \ AS^2=a^2+q^2-2aq\cos\beta \quad\to\quad b^2+h^2=a^2+a^2+b^2+h^2-2aq\cos\beta \quad\to\quad
\quad\to\quad \cos\beta=\fr{a}{q}=\fr{\sq{14}}{14}
z tw. kosinusów w  \triangle ABM\ \ MA^2=a^2+x^2-2ax\cos\beta \quad\to\quad MA^2=x^2-\fr{\sq{14}}{7}x+1
z tw. kosinusów w  \triangle DBS\ \ h^2=p^2+q^2-2pq\cos\gamma \quad\to\quad \cos\gamma=\fr{p^2+q^2-h^2}{2pq}=\fr5{\sq{70}}
z tw. kosinusów w  \triangle DBM\ \ MD^2=p^2+x^2-2px\cos\gamma \quad\to\quad MD^2=x^2-\fr{5\sq{14}}{7}x+5
f(x)=MA+MD=\sq{x^2-\fr{\sq{14}}{7}x+1}+\sq{x^2-\fr{5\sq{14}}{7}x+5}
f_{min}  jest dla  x=\fr{3\sq{910}-5\sq{14}}{112}\approx0,641